Problem 1446

1. **Checkerboard Coloring Argument for Odd $n$:**

Suppose $n$ is odd. We can color the $n \times n$ grid in a checkerboard pattern, where each square is either black or white, and adjacent squares have different colors.

- In a checkerboard pattern, the number of black squares and white squares are not equal when $n$ is odd. Specifically, there are $\frac{n^2 - 1}{2}$ squares of one color and $\frac{n^2 + 1}{2}$ squares of the other color.
- When a student moves to an adjacent chair, they move from a black square to a white square or vice versa.
- Since the number of black and white squares are unequal, it is impossible for every student to move to a new chair while maintaining the checkerboard pattern. This is because there would be an imbalance in the number of students moving to black squares versus white squares.

Therefore, it is impossible to rearrange the students according to the given rules if $n$ is odd.

2. **Constructive Argument for Even $n$:**

Suppose $n$ is even. In this case, the number of black squares and white squares in the checkerboard pattern are equal. Specifically, there are $\frac{n^2}{2}$ black squares and $\frac{n^2}{2}$ white squares.

- We can rearrange the students by having each student swap seats with the student right next to them in the same row. For example, if the students in a row are initially seated as $1, 2, 3, \ldots, 2k$, we can have:
- Student 1 swap with Student 2
- Student 3 swap with Student 4
- And so on, up to Student $2k-1$ swapping with Student $2k$.

- This ensures that every student moves to a new chair, and since $n$ is even, this swapping can be done for every row without any issues.

Therefore, it is possible to rearrange the students according to the given rules if $n$ is even.

$\blacksquare$